Permutations and Combinations Test 19

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Permutations and Combinations Test 19

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Weekly Quiz Competition

Question 1
1 / -0

If $$^nC_{10} = ^nC_{12}$$, then find n?

A
$$120$$

B
$$22$$

C
$$12$$

D
$$2$$

Solution

$$^{ n }{ C }_{ 10 }=^{ n }{ C }_{ 12 }$$
We know
$$^{ n }{ C }_{ r }=^{ n }{ C }_{ n-r }$$
$$\Rightarrow r+(n-r)=n\\ =10+12=n\\ \Rightarrow n=22$$
There fore the required value of $$n=22$$
Question 2
1 / -0

A box contains $$2$$ white balls,$$3$$ black balls and $$4$$ red balls.In how many ways can three balls can be drawn from the box if atleast one black ball is to be included in the draw?

A
$$32$$

B
$$48$$

C
$$64$$

D
$$96$$

Solution

$$2$$ White balls

$$3$$ Black balls

$$4$$ Red balls

Three balls are drawn

Total possible combination

$$\Rightarrow ^{ 9 }{ P }_{ 3 }$$

Combination without black

$$\Rightarrow ^{ 2 }{ C_{ 2 } }\times ^{ 4 }{ C }_{ 1 }+^{ 2 }{ C }_{ 1 }\times ^{ 4 }{ C }_{ 2 }+^{ 4 }{ C }_{ 3 }$$

The combination with atleast one black

$$\Rightarrow \dfrac { 9\times 8\times 7 }{ 3\times 2\times 1 } -\left[ 1\times 4+2\times \dfrac { 12 }{ 2 } +4 \right] \\ \Rightarrow 12(7)-\left[ 20 \right] \\ \Rightarrow 84-20=64$$

$$64$$ Combination include atleast one black

Hence the correct answer is $$64$$
Question 3
1 / -0

How many quadrilaterals can be formed by joining the vertices of an octagon?

A
$$65$$

B
$$60$$

C
$$70$$

D
$$64$$

Solution

Octagon Number of sides $$= n=8$$
For quadrilatral there are only four sides
There should be a selection of four ponts and arranging of quadrilatral
$$\Rightarrow ^{ 8 }{ C }_{ 4 }=\cfrac { 8! }{ 4!\times 4! } \\ \Rightarrow \cfrac { 8\times 7\times 6\times 5 }{ 4\times 3\times 2 } \\ \Rightarrow 70\quad quadrilatrals$$
Option $$: C$$
Question 4
1 / -0

The number of values of $$r$$ satisfying the equation $$\:^{69}C_{3r-1}-\:^{69}C_{r^2}=\:^{69}C_{r^2-1}-\:^{69}C_{3r}$$ is:

A
$$1$$

B
$$2$$

C
$$4$$

D
$$7$$

Solution

Solution:
$${}^{69}C_{3r-1}-{}^{69}C_{r^2}={}^{69}C_{r^2-1}-{}^{69}C_{3r}$$
or, $${}^{69}C_{3r-1}+{}^{69}C_{3r}={}^{69}C_{r^2}+{}^{69}C_{r^2-1}$$
or, $${}^{70}C_{3r}={}^{70}C_{r^2}$$
Here, $$3r=r^2$$ or $$3r+r^2=70$$ $$(\because {}^n C_r={}^n C s$$ then $$r=s$$ or $$r+s=n)$$
or, $$3r=r^2$$
or, $$r(r-3)=0$$
or, $$r=0$$ which is not possible.
or $$r=3$$
$$r^2+3r-70=0$$
or, $$(r-7)(r+10)=0$$
or, $$r=7$$ or $$r=-10$$ which is not possible.
The value of $$r$$ satisfying the equation $$=3,7$$
Number of values of $$r$$ satisfying the equation $$=2$$
Hence, B is the correct option.
Question 5
1 / -0

An event manager has ten patterns of chairs and eight patterns of tables. In how many ways can he make a pair of table and chair?

A
$$100$$

B
$$80$$

C
$$110$$

D
$$64$$

Solution

There are $$10$$ patterns of chairs and $$8$$ patterns of tables.
A pair of table and chair is required .
Therefore $$=^{10}C_1 \times 8C_1=80$$
Question 6
1 / -0

If $${ _{ }^{ n }{ C } }_{ 8 }={ _{ }^{ n }{ C } }_{ 5 }$$, then the value of $$n$$ is

A
$$2$$

B
$$3$$

C
$$1$$

D
$$13$$

Solution

Given, $${ _{ }^{ n }{ C } }_{ 8 }={ _{ }^{ n }{ C } }_{ 5 }$$.
We have the formula for combination,
$$_{ }^{n}{C}_{r}= \ _{ }^{n}{C}_{(n-r)}$$
$$\implies$$ $$n-r=5$$, if $$r=8$$ OR
$$n-r=8$$, if $$r=5$$
$$\implies$$ $$n-8=5$$ OR $$n-5=8$$
From both the equations we get $$n=13$$.
Hence, option $$D$$ is correct.
Question 7
1 / -0

When two coins are tossed and a cubical dice is rolled, then the total outcomes for the compound event is

A
$$48$$

B
$$52$$

C
$$36$$

D
$$24$$

Solution

Outcomes when $$2$$ coin tossed $$=2^2=4$$
Outcomes when dice is rolled $$=6$$
According to basic multiplication principle the total outcomes of $$n$$ independent events is calculated by multiplying outcomes of each and every individual outcomes.
$$\Rightarrow$$ Total outcome $$=4\times6=24.$$
Hence, the answer is $$24.$$
Question 8
1 / -0

Simplify: $${ _{ }^{ 34 }{ C } }_{ 5 }+\sum _{ r=0 }^{ 4 }{ { _{ }^{ (38-r) }{ C } }_{ 4 } } $$.

A
$${ }^{38}C_4$$

B
$${ }^{39}C_4$$

C
$${ }^{38}C_5$$

D
$${ }^{39}C_5$$

Solution

$$\displaystyle { }^{34}C_5 + \sum_{r = 0}^{4}$$ $${ }^{38 - r} C_4$$
$$= { }^{34}C_5 + { }^{38}C_4 + { }^{37}C_4 + { }^{36}C_4 + { }^{35}C_4 + { }^{34}C_4$$
$$= ({ }^{34}C_5 + { }^{34}C_4) + { }^{35}C_4 + { }^{36}C_4 + { }^{37}C_4 + { }^{38}C_4$$
$$= { }^{35}C_5 + { }^{35}C_4 + { }^{36}C_4 + { }^{37}C_4 + { }^{38}C_4$$ (Using $${ }^nC_r + { }^nC_{r-1} = { }^{n+1}C_r$$)
$$ = { }^{36}C_5 + { }^{36}C_4 + { }^{37}C_4 + { }^{38}C_4$$
$$ = { }^{37}C_5 + { }^{37}C_4 + { }^{38}C_4$$
$$ = { }^{38}C_5 + { }^{38}C_4$$
$$ = { }^{39}C_5$$
Question 9
1 / -0

If $$C\left( 28,2r \right) =C\left( 28,2r-4 \right) $$, then what is $$r$$ equal to?

A
$$7$$

B
$$8$$

C
$$12$$

D
$$16$$

Solution

We know that, $${}^{n}{C}_{k}={}^{n}{C}_{n-k}$$
Here $$n=28, k=2r$$
$$\Rightarrow n-k=2r-4$$ ..... $$[\because {}^{28}C_{2r}={}^{28}{C}_{2r-4}]$$
$$\Rightarrow 28-2r=2r-4$$
$$\Rightarrow 4r=32$$
$$\Rightarrow r=8$$
Hence, $$r=8$$
Question 10
1 / -0

Out of $$7$$ consonants and $$4$$ vowels, words are formed each having $$3$$ consonants and $$2$$ vowels. The number of such words that can be formed is

A
$$210$$

B
$$25200$$

C
$$2520$$

D
$$302400$$

Solution

Total no. of consonants is $$7$$ and of vowels is $$4$$
Out of $$7$$ consonants, $$3$$ are chosen to form a word. So, this can be done in $${ }^{7}C_{3}$$ ways
Out of $$4$$ vowels, $$2$$ are chosen to form a word. So, that can be done in $${ }^{4}C_{2}$$ ways
So, the no. of select $$3$$ consonants and $$2$$ vowels is $$^{7}C_{3} \times ^{4}C_{2}$$
Now, we can arrange the word containing $$5$$ letters in $$5!$$ ways.
Thus, the total no. of words formed each having $$3$$ consonants and $$2$$ vowels is
$$^{7}C_{3} \times ^{4}C_{2}\times 5! = 25200$$
Hence, the answer is $$25200$$.

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Re-Attempt Weekly Quiz Competition

Permutations and Combinations Test 19

Result

Permutations and Combinations Test 19

Score

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Rank

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SHARING IS CARING

Question 1

$$120$$

$$22$$

$$12$$

$$2$$

Solution

Question 2

$$32$$

$$48$$

$$64$$

$$96$$

Solution

Question 3

$$65$$

$$60$$

$$70$$

$$64$$

Solution

Question 4

$$1$$

$$2$$

$$4$$

$$7$$

Solution

Question 5

$$100$$

$$80$$

$$110$$

$$64$$

Solution

Question 6

$$2$$

$$3$$

$$1$$

$$13$$

Solution

Question 7

$$48$$

$$52$$

$$36$$

$$24$$

Solution

Question 8

$${ }^{38}C_4$$

$${ }^{39}C_4$$

$${ }^{38}C_5$$

$${ }^{39}C_5$$

Solution

Question 9

$$7$$

$$8$$

$$12$$

$$16$$

Solution

Question 10

$$210$$

$$25200$$

$$2520$$

$$302400$$

Solution

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